Question

If $a^2,b^2,c^2$ are in $AP$, $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{b+a}$ are also in $AP.$

• A. True
• B. False

Answer 1

The correct option is A True

If $a^2,b^2,c^2$ are in $A.P.$

$b^2-a^2=c^2-b^2$

$(b-a)(b+a)=(c-b)(b+c)$

$\frac{b-a}{b+c}=\frac{c-b}{b+a}$

$\frac{b-a}{(a+c)(b+c)}=\frac{c-b}{(a+c)(b+a)}$

$\frac{b+c-c-a}{(a+c)(b+c)}=\frac{c+a-a-b}{(a+c)(b+a)}$

$\frac{1}{a+c}-\frac{1}{b+c}=\frac{1}{b+a}-\frac{1}{a+c}$

$\therefore \frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}$ are in $A.P.$

$\frac{a+b+c}{b+c},\frac{a+b+c}{a+c},\frac{a+b+c}{a+b}$

$\frac{a}{b+c}+1,\frac{b}{a+c}+1,\frac{c}{a+b}+1$ are in A.P.

$\therefore \frac{a}{b+c},\frac{b}{a+c},\frac{c}{a+b}$ are in A.P.